Math 342
Homework 31
Daniel Walton
April 2, 2014
19.2: 2, 5, 6, 8
Exercise (2). Evaluate
z2
c2 ≤ 1}.
RRR
V
|xyz|dxdydx where V = {(x, y, z) ∈ R3 :
Solution (2). We let Ψ(x, y, z) = (ax, by, cz), then DΨ(x, y, z) =
3
2
2
2
x2
a2
a 0 0
0 b 0
0 0 c
2
+ yb2 +
, det(DΨ) =
abc. If we let K = {(u, v, w) ∈ R : u + v + w ≤ 1}, Ψ(K) = {(x, y, z) :
R
y2
x2
z2
Theorem,
|xyz|dxdydz =
V
R
Ra2 + b2 + c2 ≤ 1} =R V . By the Change of Variables
2 2 2
|uvw|dudvdw.
We now
|xyz|dxdydz
=
|aubvcw||abc|dxdydz
=
a
b
c
K
V
K
change variables into spherical coordinates as on page 580 (which shows it is a
smooth change of variables) to get
a2 b2 c2
Z
|uvw|dudvdw = a2 b2 c2
K
Z
0
2π
π
Z
0
2 2 2
1
Z
Z
|ρ3 sin2 φ cos φ sin θ cos θ||ρ2 sin φ|dρdφdθ
0
2π
π
Z
Z
1
0
2π
Z
ρ5 sin3 φ| cos φ sin θ cos θ|dρdφdθ
=a b c
0
0
2 2 2
Z
π
=a b c
0
0
1
sin3 φ| cos φ sin θ cos θ|dφdθ
6
Now do another change of variables on φ with u = sin φ, du = cos φdφ, which is
invertible on [0, π/2], and we get
a2 b2 c2
6
Z
2π
Z
1
0
Z
3
u | sin θ cos θ|du −
0
0
u | sin θ cos θ|du dθ
3
1
a2 b2 c2
=
3
=
Z
2π
Z
1
u3 | sin θ cos θ|dudθ
0
0
2 2 2 Z 2π
a b c
3
0
1
| sin θ cos θ|dθ
4
1
Now we can use a clever trigonometric identity: 2 sin θ cos θ = sin 2θ to get
Z
a2 b2 c2 2π 1
| sin 2θ|dθ
12
2
0
We’re almost there. We do something similar to above to take care of the
absolute value–split the integral into four parts, where sin has the same sign on
the interval:
#
"Z
Z 2π
Z 3π/2
Z π
π/2
a2 b2 c2
sin 2θdθ
sin 2θdθ −
sin 2θdθ +
sin 2θdθ −
24
3π/2
π
π/2
0
=
a2 b2 c2
[1 + 1 + 1 + 1]
24
a2 b2 c2
=
6
Exercise (5). For r > 0 and h > 0, show that the volume of the cone C =
{(x, y, z) ∈ R3 : x2 + y 2 ≤ r2 , 0 ≤ z ≤ h/r2 (r2 − x2 − y 2 )} is equal to 31 πr2 h.
Solution (5). We first establish that the given problem is incorrect by finding
that the true set C as a cone should actually be defined as
p
h
C = {(x, y, z) ∈ R3 : x2 + y 2 ≤ r2 , 0 ≤ z ≤ (r2 − x2 + y 2 )}
r
I don’t show this, but if you calculate the volume using the set given in the
book, you get 21 πr2 h, which is not what we’re looking for. So, using the correct
C, we know that the volume is given by the integral
Z
1dxdydz
C
so we can do a change
Ψ(ρ, θ, z) = (ρ cos θ, ρ sin θ, z) which has
cos θ −ρofsinvariables
θ 0
derivative DΨ = sin θ ρ cos θ 0 which has determinant ρ, so this is invertible,
0
0
1
and we can apply the theorem. We let (x, y, z) = (ρ cos θ, ρ sin θ, z) and get
Z
Z 2π Z r Z hr (r−ρ)
1dxdydz =
ρdzdρdθ
C
0
Z
0
0
2π Z r
h
(r − ρ)ρdρdθ
0
0 r
r
Z 2π h
1 2
hρ − ρ3 dθ
=
2
3r
0
0
Z 2π
h 2
2πhr2
=
r dθ =
6
6
0
πhr2
=
3
=
as desired.
2
Exercise (6). Suppose that Ψ : O → Rn is a smooth change of variables on
an open subset O ⊆ Rn . Use the Inverse Function Theorem to show that the
inverse mapping Ψ−1 : Ψ(O) → Rn is a smooth change of variables on the open
subset Ψ(O) of Rn .
Solution (6). Since Ψ is 1-1 and is surjective on its image Ψ(O), Ψ is a bijection
between Ψ(O) and O, so we can define an inverse Ψ−1 : Ψ(O) → Rn that is 1-1.
This shows that Ψ−1 satisfies condition (i) of a smooth change of variables. To
show condition (ii), note that Ψ is continuously differentiable and its derivative
is invertible for all x ∈ O, so it satisfies the hypotheses for the Inverse Function
Theorem, which says that D(Ψ−1 (y)) = [DΨ(x)]−1 for each point y ∈ Ψ(O),
where Ψ(x) = y, which also shows that DΨ−1 is invertible, which is condition
(ii) of a smooth change of variables.
Exercise (8). Define D = {(x, y) : x > 0, y > 0, 1 ≤ x2 − y 2 ≤ 9, 2 ≤ xy ≤ 4}.
For a continuous function
R f : D → R, use the hyperbolic coordinates from
exercise 7 to show that D [x2 + y 2 ]dxdy = 8.
Solution (8). We assume from Exercise 7 that Φ is a smooth change of variables. Start with the integral
Z
1dxdy
K
where K = {(x, y) ∈ R2 : x > 0, y > 0, 1 ≤ x ≤ 9, 2 ≤ y ≤ 4}. Also notice that
det DΦ = 2x2 + 2y 2 , so applying the change of variables we get
Z
Z
Z
2
2
1dxdy =
2[x + y ]dxdy = 2 [x2 + y 2 ]dxdy
K
so we get
R
1
2
K
D
D
[x2 + y 2 ]dxdy, and so
Z
Z
1
2
2
[x + y ]dxdy =
1dxdy
2 K
D
Z Z
1 9 4
1
=
1dxdy = (9 − 1)(4 − 2) = 8
2 1 2
2
1dxdy =
R
D
as desired.
3